Question: Simplify the following expression and state the condition under which the simplification is valid: $p = \dfrac{k^2 + k}{k^2 + 6k + 5}$
Solution: First factor the expressions in the numerator and denominator. $ \dfrac{k^2 + k}{k^2 + 6k + 5} = \dfrac{(k)(k + 1)}{(k + 5)(k + 1)} $ Notice that the term $(k + 1)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(k + 1)$ gives: $p = \dfrac{k}{k + 5}$ Since we divided by $(k + 1)$, $k \neq -1$. $p = \dfrac{k}{k + 5}; \space k \neq -1$